Minimize subject to the constraints :

Minimize subject to the constraints and :

Minimize subject to the constraints and the bounds :

Use VectorLessEqual to express several LessEqual inequality constraints at once:

Use v<= to enter the vector inequality in a compact form:

An equivalent form using scalar inequalities:

Use VectorGreaterEqual to express several GreaterEqual inequality constraints at once:

Use v>= to enter the vector inequality in a compact form:

Use Equal to express several equality constraints at once:

An equivalent form using several scalar equalities:

Specify the constraints using a combination of scalar and vector inequalities:

An equivalent form using scalar inequalities:

Minimize subject to . Use a vector variable :

An equivalent form using scalar variables:

Use vector variables and vector inequalities to specify the problem:

Minimize the function subject to the constraint :

Use Indexed to access components of a vector variable, e.g. :

Use constant parameter equations to specify the coefficients of the objective and constraints:

Use constant parameter equations to specify coefficients of several constraints:

Using vector variables:

Minimize the function subject to :

Use Vectors[n] to specify the dimension of vector variables when needed:

Specify non-negative constraints using NonNegativeReals ():

An equivalent form using vector inequalities:

Specify non-positive constraints using NonPositiveReals ():

An equivalent form using vector inequalities:

Minimize subject to by directly giving the matrix and vectors and :

The equivalent form using variables and inequalities:

Specify integer domain constraints using Integers:

Specify integer domain constraints on vector variables using Vectors[n,Integers]:

Specify non-negative integer constraints using NonNegativeIntegers ():

An equivalent form using vector inequalities:

Specify non-positive integer constraints using NonPositiveIntegers ():

An equivalent form using vector inequalities:

Specify complex variables using Complexes:

Minimize a real objective with complex variables and complex constraints :

Suppose that ; then expanding out the constraint into real components gives:

Solve the problem with real-valued objective and complex variables and constraints:

Solve the same problem with real variables and constraints:

Minimize the function subject to the constraints :

Get the primal minimizer as a vector:

Get the minimal value:

Obtain the primal minimum value using symbolic inputs:

Extract the matrix-vector inputs of the optimization problem:

Solve using the matrix-vector form:

Recover the symbolic primal value by adding the objective constant:

Find the slack associated with a minimization problem:

Get the primal minimizer and the constraint matrices:

The slack for inequality constraints is a vector such that :

The slack for the equality constraints is a vector such that :

The equality slack is typically a zero vector:

Minimize subject to and :

The dual problem is to maximize subject to and :

The primal minimum value and the dual maximum value coincide because of strong duality:

That is the same as having a duality gap of zero. In general, at optimal points:

Construct the dual problem using constraint matrices extracted from the primal problem:

Extract the objective vector and constraint matrices:

The dual problem is to maximize subject to and :

Get the dual maximum value directly using solution properties:

Get the dual maximizer directly using solution properties:

Use "ConstraintSensitivity" to find the change in optimal value due to constraint relaxations:

The first vector is inequality sensitivity and the second is equality sensitivity:

Consider new constraints where is the relaxation:

The expected new optimal value is given by:

Compare to directly solving the relaxed problem:

Each sensitivity is associated with an inequality or equality constraint:

Extract the constraints:

The inequality constraints and their associated sensitivity:

The equality constraints and their associated sensitivity:

The change in optimal value due to constraint relaxation is proportional to the sensitivity value:

Compute the minimal value and constraint sensitivity:

A zero sensitivity will not change the optimal value if the constraint is relaxed:

A negative sensitivity will decrease the optimal value:

A positive sensitivity will increase the optimal value:

The "ConstraintSensitivity" is related to the dual maximizer of the problem:

The inequality sensitivity satisfies , where is the dual inequality maximizer:

The equality sensitivity satisfies , where is the dual equality maximizer:

The default method for MachinePrecision input is "CLP":

The default method for arbitrary or infinite WorkingPrecision is "Simplex":

All methods work for MachinePrecision input:

"Simplex" and "RevisedSimplex" can be used for arbitrary- and infinite-precision inputs:

Different methods have different strengths, which is typically problem and implementation dependent:

"Simplex" and "RevisedSimplex" are good for small dense problems:

"InteriorPoint" and "CLP" are good for large sparse problems:

"InteriorPoint" or "CLP" methods may not be able to tell if a problem is infeasible or unbounded:

"Simplex" and "RevisedSimplex" can detect unbounded or infeasible problems:

Different methods may give different results for problems with more than one optimal solution:

"InteriorPoint" may return a solution in the middle of the optimal solution set:

"Simplex" always returns a solution at a corner of the optimal solution set:

A smaller Tolerance setting gives a more precise result:

Specify the constraint matrices:

Find the error between computed and exact minimum value using different Tolerance settings:

Visualize the change in minimum value error with respect to tolerance:

A smaller Tolerance setting gives a more precise answer, but typically takes longer to compute:

A smaller tolerance takes longer:

The tighter tolerance gives a more precise answer:

LinearOptimization infers the WorkingPrecision to use from the input:

MachinePrecision input gives MachinePrecision output:

LinearOptimization can compute results using higher working precision:

If the specified precision is less than the precision of the input arguments, a message is issued:

Maximize subject to . Solve a maximization problem by negating the objective function:

Negate the primal minimum value to get the corresponding maximal value:

Convert into a linear function using and :

Convert into a linear function using and :

Minimize subject to the constraints . Convert into a linear function using :

The problem can also be converted using :

Minimize subject to the constraints . The constraint can be transformed to :

Minimize subject to the constraints . The constraint can be interpreted as . Solve the problem with each constraint:

The optimal solution is the minimum of the two solutions:

Minimize subject to , where is a nondecreasing function, by instead minimizing . The primal minimizer will remain the same for both problems. Consider minimizing subject to :

The true minimum value can be obtained by applying the function to the primal minimum value:

Minimize , subject to :

Since , solve the problem with :

The true minimum value can be obtained by applying the function to the primal minimum value:

Find a robust linear fit to discrete data by minimizing . Using auxiliary variables , the problem is transformed to minimize subject to the constraints , which is a linear optimization problem:

Fit the data:

A plot of the line shows that it is robust to the outliers:

Compare with the fit obtained using LeastSquares:

Find a linear fit by minimizing such that the line should pass through the first data point:

By adding the constraint a〚1〛.xb〚1〛, the line is forced to pass through the first point:

The plot shows that the line passes through the first data point while still maintaining a robust fit:

Find a robust fit to nonlinear discrete data by minimizing . Generate some noisy data:

Fit the data using the bases . The approximating function will be :

Find the coefficients of the approximating function:

Visualize the fit:

Compare the approximating model with the reference and the least squares fit:

Find a linear fit to discrete data by minimizing . Since , by introducing the scalar auxiliary variable , the problem can be transformed to minimize , subject to the constraints :

Fit the data:

The line represents the "best-worst case" fit:

The function Fit can be used directly to compute the coefficients:

Find a line that separates two groups of points and :

For separation, set 1 must satisfy and set 2 must satisfy :

The separating line is:

Find a quadratic polynomial that separates two groups of points and :

For separation, set 1 must satisfy and set 2 must satisfy :

The coefficients of the quadratic polynomial are:

The quadratic polynomial is:

Find the minimum-degree polynomial that can separate two sets of points in the plane:

Define a polynomial power function that avoids issues when or is 0:

Define a function that gives a polynomial of degree with coefficient as a function of :

The variables for a degree are the coefficients:

The constraints enforce separation between set 1 and set 2:

For example, the constraints for quadratics are:

For separation, the only condition is that all the constraints must be satisfied. The objective vector is set to 0:

Find a separating polynomial and show it with the points:

Find the mix of products to manufacture that maximizes the profits of a company. The company makes three products. The revenue per unit, cost per unit and maximum capacity are given by:

Each product is made using a combination of four machines. The time a machine spends on each product is:

The profit equals revenue minus cost times the number of units for product :

Each machine can run at most 2400 minutes per week:

The company makes between zero and the capacity for each product:

The optimal product mix is given by:

The resulting profit is given by:

The breakdown of machining time for each product is:

The same company wants to find out what capacity to change that will have the largest impact on profit. The optimal product mix and profit as formulated previously are given by:

The profit based on the current capacity is:

Extract all the data needed to analyze the impact of the capacity change on profit:

The sensitivity gives the derivative of the profit wrt each of the constraints. Extract the constraints that have any sensitivity at all:

Extract the sensitive constraints from :

If the production capacity for product 3 is increased, there should be an increase in profit:

The resulting increase in profit:

This corresponds exactly to the predicted increase from the sensitivity:

If the production capacity of product 1, product 2 or both is changed, the profit does not increase:

Find the mix of products to manufacture that maximizes the profits of a company. The company makes three products . The revenue per unit is:

The factory is only open for 100 days. It takes days, respectively, to make each unit:

Each product contains gold and tin. The total gold and tin available are 30 and 200 units, respectively:

The company incurs a setup cost for each product if that product is manufactured:

Because of the relatively high setup costs, it may be desirable not to manufacture some of the products at all. Let be a decision variable such that if product is manufactured and otherwise . The limit ensures that if is 0, so is , without limiting the size of more than other constraints do:

The objective is to maximize the profit:

The optimal mix is given by:

John Conner has learned that Skynet is planning on attacking five settlements. He needs to deploy soldiers from three base camps to the settlements to meet the threat. The fuel cost to transport one soldier from base camp to settlement is:

Let represent the number of soldiers to be shipped from base camp to settlement . The cost function to minimize is:

Each base camp needs to distribute a maximum of 100 soldiers between the settlements:

Each settlement needs at least 50 soldiers from a combination of the base camps:

The number of soldiers to be deployed from each base camp to each settlement is:

The breakdown of soldier deployment from each base camp is:

The Jedi Order is fighting the separatist armies led by Dooku. Dooku and two of his generals, Asajj Ventress and General Grievous, have attacked three outer colonies. The Jedi Order sends three Jedi to assist the outer colonies. Let be the odds of Jedi beating Sith :

Let be a matrix variable where is 1 if Jedi battles Sith , and 0 otherwise. The cost function is the expected number of defeats given by and constructed using Inactive[Times]:

Since the outer colonies are far apart, each Jedi may only be assigned to one Sith, and vice versa. Also, since is 0 or 1, then :

Find the appropriate Jedi to face the Sith to minimize the chances of defeat:

This means that Obi-Wan Kenobi is sent to General Grievous, Mace Windu is sent to Asajj Ventress and Ki-Adi-Mundi is sent to Dooku. The expected number of defeats is:

Solve a bipartite matching problem between two sets and that minimizes the total weight:

Let be a matrix variable, where is 1 when is matched to and 0 otherwise. The total weight of the matching is given by and constructed using Inactive[Times]:

The row and column sums of need to be 1 and also :

Use LinearOptimization to solve the problem:

The matching permutation may be extracted by converting the matrix into a graph. Round is used to truncate small values that may occur due to numerical error in some methods:

Use the adjacency graph to build a bipartite graph: